Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $n = \dfrac{t - 10}{4t^2 - 8t - 320} \div \dfrac{t^2 - t}{t^3 + 5t^2 - 6t} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{t - 10}{4t^2 - 8t - 320} \times \dfrac{t^3 + 5t^2 - 6t}{t^2 - t} $ First factor out any common factors. $n = \dfrac{t - 10}{4(t^2 - 2t - 80)} \times \dfrac{t(t^2 + 5t - 6)}{t(t - 1)} $ Then factor the quadratic expressions. $n = \dfrac {t - 10} {4(t - 10)(t + 8)} \times \dfrac {t(t - 1)(t + 6)} {t(t - 1)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {(t - 10) \times t(t - 1)(t + 6) } { 4(t - 10)(t + 8) \times t(t - 1)} $ $n = \dfrac {t(t - 1)(t + 6)(t - 10)} {4t(t - 10)(t + 8)(t - 1)} $ Notice that $(t - 10)$ and $(t - 1)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {t(t - 1)(t + 6)\cancel{(t - 10)}} {4t\cancel{(t - 10)}(t + 8)(t - 1)} $ We are dividing by $t - 10$ , so $t - 10 \neq 0$ Therefore, $t \neq 10$ $n = \dfrac {t\cancel{(t - 1)}(t + 6)\cancel{(t - 10)}} {4t\cancel{(t - 10)}(t + 8)\cancel{(t - 1)}} $ We are dividing by $t - 1$ , so $t - 1 \neq 0$ Therefore, $t \neq 1$ $n = \dfrac {t(t + 6)} {4t(t + 8)} $ $ n = \dfrac{t + 6}{4(t + 8)}; t \neq 10; t \neq 1 $